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3 years ago

Plz plz helpppppp0pp

Mathematics

1 answer:

postnew [5]3 years ago

5 0

Answer:

28cm

Step-by-step explanation:

Hi!

Let's find the area of the rectangle (we can form) and then tack on the area of the triangles.

We have 4 * 3 = 12. So the first area is 12.

Now, the triangles. We see that they will have base 4, and leg 2.

Using the Pythagorean Theorem:

4² + 2² = 20. √20 = 2√5. That won't really matter however, because we will multiply 4 by 2 to find the area.

4 * 2 = 8

The same will happen on the other side:

4 * 2 = 8 (note that we could've created a rectangle out of the triangles, but that would give us 16 too.)

8 + 8 = 16.

16 + 12 = 28

Thus, the area is 28 cm.

Hope this helps!

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3 years ago

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Step-by-step explanation:

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3 years ago

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Sedaia [141]

Complete Question

Answer:

$SE = 0.66}$

$-3.29 < \mu_1 - \mu_2 < -0.70$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The first sample mean is $\= x _1 = 8$

The second sample mean is $\= x _2 = 10$

The first variance is $v_1 = 0.25$

The first variance is $v_2 = 0.55$

Given that the confidence level is 95% then the level of significance is 5% = 0.05

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the first standard deviation is

$\sigma_1 = \sqrt{v_1}$

=> $\sigma_1 = \sqrt{0.25}$

=> $\sigma_1 = 0.5$

Generally the second standard deviation is

$\sigma_2 = \sqrt{v_2}$

=> $\sigma_2 = \sqrt{0.55}$

=> $\sigma_2 = 0.742$

Generally the first standard error is

$SE_1 = \frac{\sigma_1}{\sqrt{n} }$

$SE_1 = \frac{0.5}{\sqrt{60} }$

$SE_1 = 0.06$

Generally the second standard error is

$SE_2 = \frac{\sigma_2}{\sqrt{n} }$

$SE_2 = \frac{0.742}{\sqrt{60} }$

$SE_2 = 0.09$

Generally the standard error of the difference between their mean scores is mathematically represented as

$SE = \sqrt{SE_1^2 + SE_2^2 }$

=> $SE = \sqrt{0.06^2 +0.09^2 }$

=> $SE = 0.66}$

Generally 95% confidence interval is mathematically represented as

$(\= x_1 -\= x_2) -(Z_{\frac{\alpha }{2} } * SE) < \mu_1 - \mu_2 < (\= x_1 -\= x_2) +(Z_{\frac{\alpha }{2} } * SE)$

=> $(8 -10) -(1.96 * 0.66) < \mu_1 - \mu_2 < (8-10) +(Z_{\frac{\alpha }{2} } * 0.66)$

=> $-3.29 < \mu_1 - \mu_2 < -0.70$

5 0

3 years ago

Help please!!! Thank you

Elena L [17]

<u>Given</u>:

The given expression is $\frac{4 d+28}{12 d+96} \cdot \frac{d^{2}+14 d+48}{d^{2}+9 d+14}$

We need to multiply the terms.

<u>Multiplication of the terms:</u>

Before multiplying the terms, first we shall find the factors of the quadratic equations.

Thus, we have;

$\frac{4 d+28}{12 d+96} \cdot \frac{(d+6)(d+8)}{(d+2)(d+7)}$

Factor out the common terms, we get;

$\frac{4 (d+7)}{12 (d+8)} \cdot \frac{(d+6)(d+8)}{(d+2)(d+7)}$

Let us cancel the common terms from the above expression.

Thus, we have;

$\frac{4 }{12 } \cdot \frac{(d+6)}{(d+2)}$

Simplifying, the term, we get;

$\frac{1}{3 } \cdot \frac{(d+6)}{(d+2)}$

Now, we shall multiply the terms.

Hence, multiplying the terms, we get;

$\frac{(d+6)}{3(d+2)}$

Thus, the multiplied value of the given expression is $\frac{(d+6)}{3(d+2)}$

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3 years ago